Find f^-1(x) <-- inverse if f(x) = x^2 - 4x + 3, x <= 2 find value of f^-1(8).
I got x=29 BUT x<= 2????
Not quite. Your original function was $\displaystyle f(x)=x^2-4x+3$
$\displaystyle f(x)=y=x^2-4x+3$
Interchange x and y
$\displaystyle x=y^2-4y+3$
Isolate the y
$\displaystyle y^2-4y=x-3$
Complete the square on the left side. You'll see why in a minute.
$\displaystyle y^2-4y+4=x-3+4$
$\displaystyle (y-2)^2=x+1$
$\displaystyle y-2=\pm\sqrt{x+1}$
$\displaystyle y=\pm\sqrt{x+1}+2$
$\displaystyle \boxed{f^{-1}(x)=\pm\sqrt{x+1}+2}$